Are Velocity and Acceleration Vectors Perpendicular When Speed Is Constant?

[jumboopizza]

Homework Statement

the question is simply asking to prove that if speed is constant than velocity and acceleration vectors are perpendicular to each other. it also says as hint to differentiate v dot v= u^2...
V=velocity A=acceleration u=speed t=time

Homework Equations

i suppose knowing the vector dot product properties would be useful to have around,so here's some:
V dot V= [v]^2(magnitude)
derivative of vector dot products:
d(a dot b)/du= da/du dot b+ a.db/du
acceleration is= the derivative of Velocity with respect to time

The Attempt at a Solution

ok so I've used the hint and tried to differentiate V dot V= u^2 and i get

dv/dt dot V+V dot dv/dt= du/dt u^2= 0 since speed is constant
2V dot dv/dt=0
now I am not sure if this is representinvg acceleration or not the 2v dv/dt or if its the rate of change in speed or whatever, if its acceleration then wouldn't it be 2A=0, but now it doesn't make sense since acceleration can't be 0 so i know there's a mistake somewhere...

my other idea was to use V dot V=[V]^2 which is one of the dot product properties
and then the magnitude of the velocity would be the speed wouldn't it?

dv/dt [V]^2=0

anyways I am lost,so can someone please help?

 

[gabbagabbahey]

Homework Statement

the question is simply asking to prove that if speed is constant than velocity and acceleration vectors are perpendicular to each other. it also says as hint to differentiate v dot v= u^2...
V=velocity A=acceleration u=speed t=time

Homework Equations

i suppose knowing the vector dot product properties would be useful to have around,so here's some:
V dot V= [v]^2(magnitude)
derivative of vector dot products:
d(a dot b)/du= da/du dot b+ a.db/du
acceleration is= the derivative of Velocity with respect to time

The Attempt at a Solution

ok so I've used the hint and tried to differentiate V dot V= u^2 and i get

dv/dt dot V+V dot dv/dt= du/dt u^2= 0 since speed is constant
2V dot dv/dt=0
now I am not sure if this is representinvg acceleration or not the 2v dv/dt or if its the rate of change in speed or whatever, if its acceleration then wouldn't it be 2A=0, but now it doesn't make sense since acceleration can't be 0 so i know there's a mistake somewhere...

my other idea was to use V dot V=[V]^2 which is one of the dot product properties
and then the magnitude of the velocity would be the speed wouldn't it?

dv/dt [V]^2=0

anyways I am lost,so can someone please help?

acceleration (a vector!) is the time derivative (rate of change) of the velocity vector, so $\mathbf{v} \cdot \frac{d \mathbf{v} }{ dt } = \mathbf{v} \cdot \mathbf{a} = 0$

Under what conditions will two vectors have a dot product of zero?

 

[jumboopizza]

ok that's what i thought dv/dt= A but its not necessarily 0, if they( vector V and Vecotor A) are perpendicular to eachoter the dot product would be zero...

ok so V dot A=0
so when i first derived V dot V= u^2 and got to dv/dt V+V dv/dt= u^2 du/dt=0

du/dt is the rate of change of speed correct? so if the first derivative of u is =0
would't u'= equal zero?

V dot A= uu' -------------> V dot A= u(0)?

 

[jumboopizza]

or is it better to make V dot V= [V]^2

sqrt [V]^2= sqrt u^2 ---------> [v]= u

d[v]/dt [v]= du/dt u = 0 --------> d[v]/dt = [a] magnitude of the acceleration

 

[gabbagabbahey]

It is better to use $v^2= |\mathbf{v}|^2$ than $u^2$, since $\mathbf{v}\cdot\mathbf{v} = |\mathbf{v}|^2$ follows from the definition of dot product, and by definition, the speed is the magnitude of the velocity.